Thursday, December 6, 2012

Remember in class we talked about "adding the opposite" whenever you see a subtraction sign. We called it "add op." For example, instead of solving the problem 32-78, we should look at it as 32 + (-78) or the problem 16 - (-32) as 16 + 32. The reason why this works has to do with the direction you are going on a number line. Subtracting (-2) means you are going to move two places to the right on the number line which is the same as adding positive 2.

The reason I suggest "add op" when you see subtraction because of two primary reasons.

When the work becomes more complicated with variables and various grouping symbols and you are expected to simpliy or solve, it becomes really easy to "lose the sign" meaning you have forgetten to include a negative sign as you are solving the problem. "Add op" helps you to prevent this common error.

"Add op" allows you to make your problem/expression into all addition. When you have all addition you can use the communitive and associative properties to group the work in ways that save time and allow you to do some mental math - essentially making the computation easier.

So you only have to worry about adding integers of the same sign or integers of opposite signs. Adding integers of the same sign is simple if you think about positive and negative counters. For example (-12) + (-13) can be viewed as 12 negative counters plus 13 negative counters, so when you add all the negative counters you will have 25 negative counters or (-25).

So we can create a rule for adding integers of the same sign. RULE: Adding integers of the same sign will result in the sum with the sign of the addends. 3+4=7 OR (-3) + (-4) = (-7)

Adding integers with opposite signs becomes a little trickier. Below are two videos that might help you understand. First is a video using a chip model. The second video is from Khan Academy using a number line.

The second rule we can create then is RULE: When adding integers of opposite signs, subtract the absolute value of the addends (the numbers you are adding). Give the difference the sign of the larger absolute value. I like to think of it as, "Do I have more negatives or more positives in my problem? If I have more negatives, my answer is negative. If I have more positives, my answer is positive."

Tuesday, November 27, 2012

We will be talking about three transformations of figures on the coordinate plane. (you will not be responsible for these concepts on any unit test. However, you may say this on an upcoming quiz or possible the NJASK)

Translation - a translation is simply a sliding of the figure to a different position on the coordinate plane. THe reason why it is called a translation is because you are taking all the x-coordinates of the original figure and moving them left or right by the same number of spaces. The same is true for the y-coordinates.

Example: The coordinates (-7,-2), (-7,-6), and (-4,-6) form a right triangle in Quadrant III. You can translate this triangle to Quandrant I by adding 9 (in this case) to all your x-coordinates and 10 to all your y-coordinates. So your translated points will be (2, 8), (2, 4) and (4, 4). Try it on grid paper to have a look.

To describe a translation you write T(x,y). For the translation below, sliding from the red triangle to the blue, we moved five spaces to the left, so we subtracted 5 (or -5). We also moved up 3 spaces. Because the x axis is left to oright and the y is up and down, the translation can be written as T(-5,3).

Reflection - a reflection is a mirror image across either the x-axis or the y-axis. Since the y-axis measures the positions of the horitzontal lines, reflecting the image across this axis means that all the y-coordinates remain the same but the oppiste of the x-coordinates are needed to reflect the figure. The same is true for reflecting across the x-axis. However, all the x-coordinates remain the same, but the y's change.

For the reflection above, notice that point A is (1,7). However the reflection or A' is (-1,7). Point B is (2,3) so the reflection across the y-axis is (-2,3). If we were to reflect the image across the x-axis, point A (1,7) transforms into point A' (1,-7).

Rotation - in regards to rotation the figure is turned a certain number of degrees around a specific point. This point is call the center of rotation. The center of rotation can be the origin or even a point one the figure.

Monday, October 1, 2012

If you missed Back to School Night or would like to see the screencast again, it is posted below. The sound quality is not the best, so the a copy of the students' explanations are below as well.

Hello everyone, this is
_______________________________ and I am here to introduce our math
website. On the site you’ll be see
important links to the left. On the home
page here you will see a link to our math program which is called connected
math. If you click on the link here for
phschool.com and type in the web-code AMK 5500, you will be navigated to many
of the connected math resources for sixth grade. Click on the features for students. Here you’ll see video tutors which are also
available on Mr. Lattanzi’s website as well as links for multiple choice
practice and vocabulary.

Hi everyone this
is_____________________ and if you look to the left under Units, you’ll see
several of the units we will complete this year. For example, if you click on our first unit
Prime Time, you will be brought to a page that has many resources for that
unit. The parent letter here is a PDF
that describes the unit but also provides a good study guide for us. There are also videos that are provided by
the phschool website. And, if you scroll
down there are PDFs of all the ACE problems in our textbook so we don’t have to
take it home. There are different types
of problems for practice. You’ll see ACE
problems, additional practice sheets, and skills sheets all by
investigation. The answers to all these
problems are also provided.

Hi everyone,
I’m__________________I’d like to show you some of the other features of the
website. All of them are available on
the menus to the left of the page. If
you would like more information about homework click on the link below “About
the Class.” Here there is an explanation
about the expectations for homework.
Briefly, homework should not take longer than thirty minutes. If problems arise simply email or send in a
note to Mr Lattanzi to explain any issues.
We students are expected to attempt all assigned problems, guessing if
necessary or explaining our thinking.
Homework is not graded but checked for completion and reviewed the next
day. You can see additional information
about grading, our math program, and Mr. Lattanzi under About the Class.

Hi everyone,
I’m___________________ I will show you a couple of our projects this year. First I’ll show you our Dream Home Project
where we’ll design a digital and hand-drawn version of the perfect home to
better understand area, perimeter, scale, and cost of materials. All of the important documents are listed on
the site as well as past student work, and videos to learn how to use small
blue printer.com

Another project we’ll complete will use our checking
account balances as a down payment to purchase a new car. We’ll complete this webquest learning new
vocabulary, figuring out our budgets, determining interest, fuel costs, and loan
payments. We’ll then share our financial
findings on a class wiki.

Hi everyone,
I’m_________________ and I’d like to show you the class blog. The blog may be the most helpful resource for
us this year. Mr. Lattanzi tries to
update it at least once a week. You’ll
see explanations about various concepts, videos, sample problems, study guides
and suggestions, and maybe even excellent student work. Students are encouraged to create their own
blog entries on word documents or create videos and images on various math
concepts.

Most of us have the website and the blog bookmarked at
home, if not we encourage you to do so or you can simply google Lattanzi math
to find the site. Thanks for listening
and good night everyone!

Wednesday, September 19, 2012

Students in periods 1,2,7, & 9 used created models as a problem solving technique to solve the complex locker problem. The directions and parameters for the problem can be found here. The following are reflections written by the students after they managed to solve the problem and understand its patterns.

Reflection of Locker Problems

By: MC, VC, CC

Our group made a model of the locker problem and we determined, using our model, (the chalupa2 method), that every square number is open because it has an odd number of factors. Sometimes, we made mistakes. Some of our mistakes were misplacing things. This shook us off course and it just took it longer to figure out the answer. We thought of the method we used in our heads and we modified it on paper. The chalupa2 method is using a 30 by 30 grid and filling out each of the boxes with symbols of open and or closed lockers. We learned that this model took a long time and that we could have done it another way. Our teacher, Mr.Lattanzi, showed us that we didn’t have to fill out the whole model to figure out the answer. Even though it took a long time to fill out, it helped us a lot because we could see what we were thinking in our head. Using this model we could see a pattern. This pattern was critical to this problem. With the pattern we could figure out the answer 10 times faster than usual. This experience helped a lot because if we were in a situation like this problem we could use the same method to figure out the answer easily.

As you can see, we had struggles with this problem. Combining our brains together helped out a lot. Our model amazed us with its success. Now, if we ever get stuck in this situation again, we can use many methods we used for this problem to help. (This model is not a patented model. People can use this to figure out problems similar to this but keep in mind that we made it.)

Locker Problem

While doing our locker project, we learned a lot. JT AR, and I (BE) first created a model that didn’t quite work the way we wanted it to. In the model, we listed all the lockers from 1 to 30 and listed all the factors for each locker. Soon we noticed that this model was disorganized. Then when our class shared some of our models, we thought another group’s idea was more organized and well-developed. That model showed us the pattern clearly between the students that were opening and closing the lockers. It included writing out the numbers 1-30 twice (once down the y-axis and the other across the x-axis) to show both the students and the lockers so we could figure out the pattern for the 1,000 students and lockers. As our project came to an end, we color coded the graph paper that we did our project on. The project helped us learn to work better with a group and taught us about divisible numbers, factors, and products. We learned the importance of a model and finding patterns when you have large numbers because it makes it simpler to figure out the problem. It is also an organized way to see all the work you did and learned from it too figure out other problems that are similar. Overall, we found out that all of the square numbers had an odd number of factors which keeps them open at the end. We learned a lot during this locker experience.

Reflection

We were assigned to do a locker problem and make a model to show our work. The locker problem is where we have to find out which lockers are open or closed at the end. We also had to find out which student touched which locker. We started off by labeling lockers and students with numbers in a 30 by 26 grid formation. Next, we started filling out the graph by putting an O and highlighting it orange for the locker being open and a C which was highlighted blue if the locker was closed. After that, we determined which student touched each locker and we searched for important patterns. Modeling is important because it helps show what we are thinking in our head, and it helps us find patterns.

The challenge started when we had to figure out how to setup and label the graph. That was a problem because if we counted wrong then we wouldn’t see the pattern. One mistake we made was we counted wrong and ended up with lots of problems. We eventually found out that all the lockers with their number being a square number (1, 4, 9, 16, 25 etc.) were opened and all the other numbered lockers were closed. What we would have done differently is we would have paid more attention to the number of lockers and students and the type of number they were. We learned that working together is better than working alone because we can get more work done and use our time wisely.

Written by: TT, EW, CB, KC

Reflection

by: AL, CB, AS, HK

In the beginning, we read the locker problem. Then, we made a small chart of thirty lockers out of a thousand to help us solve it. Our goal was to see which lockers would be closed or open after the one thousandth locker and who touched which locker. Modeling was an important part because it helped us visualize the problem before we solved it. The challenges we had in this problem was figuring out the pattern with the square numbers. An occasional mistake we made was getting open and closed lockers confused. If we had the chance to do it again we would have done the color coding and labeling differently. This experience would help us when using combinations and problem solving. Patterns are an important part because they would help us if we had to figure out bigger numbers that weren’t on our chart. As a group, we enjoyed drawing the model and figuring out the problem which helped us learn when we have a big number problem to find the pattern.

Locker Reflection - BS, LH, NKOur group did a project about who opened what locker. We decided that we couldn't do all the math in our head so we made a chart on graph paper. We went across and down making lines 1-30. Person 1 opened all the lockers. Person 2 closed every other locker. Person 3 changed every 3rd locker, so say that 3 was closed and 6 was open. person 3 would open 3 and close 6 but leave 1,2,4, and 5 the same way they had found it. So person 4 changed every 4th locker, or every multiple of 4 because they were the 4th person, so the number of the person, the multiple of the locker they change. In our chart, we put c’s for closed and o’s for open. Lots of times we mixed up the letters and realized that the whole thing was wrong. We had to start over. On diagonal all of the lockers were changed. We looked at the real pattern and found that on diagonal 1,4,9,16, and 25 were open. Do you know why they were open? Because they are all square numbers! One of the mistakes we made is when we finished, we traced it with marker and realized it was wrong, so next time we should check our work more carefully. This project helped us realize that we should organize our thoughts better by using graphs instead trying to figure it out in our heads. We have seen that car companies use models of cars that they want to sell before making the real version. They can see how the cars would look before creating it. A pattern is important because you can see how everything is related to each other and it makes it easier to solve problems with bigger numbers. There were definitely challenges involved in the project. One of the challenges was that in the beginning we couldn't find what kind of model to use to figure out the problem. Another challenge is that also in the beginning, some of us didn't know how to solve the problem, but then we all helped each other and came up with a solution. In the end, the locker problem was a fun, hard working project and we enjoyed trying to figure it out.

Tuesday, September 18, 2012

Hi Guys, Remember this from class? We talked about why this method is useful and how it keeps us organized, but one of the great things about this method is that if you are stuck or can't think of a factor pair quickly, you can skip it and come back later using the other factors you found as clues.

Monday, March 26, 2012

Here is an excellent video made by one of your classmates to help you understand how to find whole number side lengths of possible rectangles given a fixed area. Remember, we use the word 'fixed' here to mean constant or "stays the same."

Fixed Area

By: MJ & VM

Fixed Area is where you find different lengths and width(using whole numbers) of rectangles with a certain area. When solving a problem about finding area, simply find factors of the area. The U turn method could be very helpful when finding factors. YOU WILL HAVE TO FIND ALL POSSIBLE DIMENSIONS! Also, perimeter of the rectangle is required. You must find the length and width before finding the perimeter. A fixed area with dimensions that is the most square like will have the smallest perimeter. The fixed area with dimensions that is the least square like will have the biggest perimeter. I believe this works just because of the squares and rectangles property’s.

Here’s how a problem would look like step by step:Find the all the possibilities of rectangles with a fixed area of 8cm2.

1. First, find each possible length and width by finding factors of eight using the U turn method.

U Turn Method

1 * 8

2 * 4

4 * 2

8 * 1

MAKE SURE YOU ALWAYS USE THE OPPOSITE OF EACH RECTANGLE!!!!!!(If a rectangle is 1 by 8, its opposite would be 8 by 1)

2. After you found all of the fixed area’s factors and its opposites, copy them into the length and width columns in a chart similar to the one below.

Chart

Length

Width

Perimeter

Area

1cm

8cm

8cm2

2cm

4cm

8cm2

4cm

2cm

8cm2

8cm

1cm

8cm2

ALWAYS LABEL YOUR MEASUREMENTS! 3. Once you have the length and width copied into the chart, find the perimeter of each rectangle. The perimeter can be determined by using the formula,2(L+W). Imagining the rectangle could help when finding perimeter.

Length

Width

Perimeter

Area

1cm

8cm

18cm

8cm2

2cm

4cm

12cm

8cm2

4cm

2cm

12cm

8cm2

8cm

1cm

18cm

8cm2

Notice how the perimeter changes. The perimeter of a shape will always stay the same but you can change the area.

Perimeter has to be involved in a fixed area problem because you need both the area and perimeter to describe size. One doesn't give enough information.

STILL REMEMBER TO LABEL PERIMETER AS WELL AS LENGTH, WIDTH, AND AREA!4. Always check over your work in case you’ve made a mistake.

Creating nets can be challenging, but if you take some time to think about the number of sides of your prism and the which lengths meet, should be able to take the three dimensional prism to create a the two dimensional net. The video video gives a clear presentation of this.

When you first take a look at the irregular rectangular shape below, you might be unsure how to find the area for the modified shape. However, finding the area for an irregular shape like the one below takes some steps in a process, but there are two different techniques to use to help find the area. First, let's take a glance at our irregular shape below.

IMPORTANT PRECAUTIONS: When you see the irregular shape, be sure to label the missing measurements so you can go on with either of the two techniques. For the width, we know that one part of the width is 5 units, and another unlabeled width line. To find the length of this line, we must subtract 7-5. So that answer is two units. So that line is 2 units. For the length, one part of the length on the right is 3 units. So we must subtract 5-3 to find the length. So that line is 2 units long. Now, our irregular shape has all the lines labeled. This can be done because opposite sides of a rectangle are equal. Since this shape can fit inside a rectangle and all lines are horizontal and vertical and create right angles, you can determine missing side lengths by subtracting the known lengths.

So now that all the lines are labeled, it's time to use either of the two ways to find our area.

WAY #1: The first way to find the area of an irregular shape is to divide it into multiple rectangles. This will make finding the area much simpler, since you will be finding the area of each divided rectangle and adding up all the areas of each of those rectangles to find the total area.

Now that the shape has been divided into smaller rectangles, let's find the area of each of the rectangles. For Shape 1: 5 X 5 = 25 units squared. For Shape 2: 2 X 3 = 6 units squared. So 25 + 6 units squared gives us our total area of 31 square units. Be sure you are using the appropriate lengths when you are finding area. Notice to find the area of Rectangle 1 we are multiplying 5 X 5 not 5 X 7 even though the length of the whole side is 7 units. That length is not the length of the rectangle we broke out of our entire figure.

Now, for the second way to find the area.

WAY #2: The other way is to make the modified shape appear as a rectangle. So, let's go ahead and make the shape into an actual rectangle.

So now, we multiply 7 X 5 to get 35 units squared for our full rectangle. After that, we will now find the area of the rectangle that was the empty space. Since we know the length and width of the rectangle, we multiply 2 X 2 to get 4 units squared. Now, we subtract the area of the full rectangle from the empty space that was made a rectangle. 35 - 4 = 31 square units, which was the answer we got from the other way to fin d the area! Using either of these processes is a matter of preference, and both will work.

JUST BE SURE TO:

TO REMEMBER TO LABEL ALL THE LINES IN THE SHAPE BEFORE PROCEEDING WITH THESE STEPS.

MAKE SURE YOU KNOW THE AREA OF THE DIVIDED RECTANGLE YOU ARE ADDING/SUBTRACTING.

Hope you enjoyed the article (and found it helpful) on finding the area for the irregular rectangular shape.

Definition- A fixed area is when you are given a number and that is your defined area. That area stays the same for all rectangles with whole number dimensions.

When finding the area of a rectangle you use the formula A=bh or Area = base * height. A fixed area is when you are given a number as the defined area. You then need to find the base and height that would give you that area. When solving for a fixed area of a rectangle you are finding factors of the fixed area to determine the dimensions of the rectangles. The factors you are looking for would be the base and height of the rectangle that, when multiplied together, give you the fixed area. Drawing a table like the one below can help you solve the problem:

Base

Height

Perimeter

Area

1 in.

16 in.

34 in.

16 square in.

2 in.

8 in.

20 in.

16 square in.

4 in.

4 in.

16 in.

16 square in.

You can see the area stays the same above and that is why it is called "fixed."

Filling the table (above) is easy. First, you make the table and add the columns and rows. Don't forget to label the columns (Base, Height, Perimeter, and Area). Next, find the factors for the fixed area and insert these into the base and height columns in pairs. Also, you must find the perimeter for a rectangle that has the base and height of the factors in the first two columns. For example, the rectangle on the first line above has a base of 1 inch and a height of 16 inches, which are two factors of 16 sq. in, the fixed area in this case. The perimeter of that rectangle would be 16+16+1+1 or 34 inches. Lastly, you have the area column to fill in, and obviously it's the same every row if there is a fixed area.

While making the table, remember to include the unit of measurement. If the area is given in square inches, then the base and height and perimeter will be in inches. If you look at the table above you, can see all the units are inches. Only the area's unit of measurement is squared.

Here are some other examples to prove that finding factors is a trick for fixed area.

You can see that 12 and 3 are factors of 36. When you multiply them you get 36. Here is a factor rainbow that shows the common factors of 12.

Since we have the factors of 12, we will use them to build a similar table with a fixed area of 12 square inches. First fill in the area (in red). Then use the common factors in the factor rainbow to fill in the base and height (in blue). Lastly, add 2b+2h to get the perimeter.

Base

Height

Perimeter

Area

1 in.

12 in.

26 in.

12 square in.

2 in.

6 in.

16 in.

12 square in.

3 in.

4 in.

14 in.

12 square in.

In summary, using factors as the base and height for a given fixed area will allow you to master this area of math.

Friday, March 2, 2012

Volume is the amount of 3-D space that a object takes up. Volume is measured in cubic units, because volume is 3-D and a cube is a 2-D square with another dimension. A rectangular prism is just what it sounds like, a prism made of rectangles. Think of a cube, but rectangular faces, so a little stretched out. A triangular prism is almost like a rectangular prism, but it’s end faces are triangles. If you look at the prisms below, you can see that there are three parts labelled h, b, and l or w. That h, b, and l or w stands for height, base and length or width (height, width and length being the first 3 dimensions, and the dimensions of a cube). So, volume is basically figuring out how many of those cubed units fit in a figure. If area is figuring out how many squared units are in a figure, and the formula for finding area for a rectangle is length by width, and a rectangular prism is a rectangle with height, it is logical for volume=height*length*width to be the formula for a rectangular prism. You may see this formula, but it is more likely that you will see v=hlw, the shortened version. The same theory goes for triangular prisms. If the formula for area is a= 1/2 lw, add height to make it v= 1/2 hlw. Below are so practice problems and videos, and a very short Voki.

The following video gives and excellent representation of how cubic units work to determine volume. Be sure to watch it to get a clear understanding.

This video explains triangular prisms and provides a couple of examples.